Question Number 156248 by alcohol last updated on 09/Oct/21
Answered by physicstutes last updated on 09/Oct/21
![e^(iθ) = cos θ + i sinθ.......(i) e^(−iθ) = cos θ − i sin θ.....(ii) equation (i) and (ii) are the polar and index (euler) form of a general complex number with modulus 1. (i) + (ii) ⇒ e^(iθ) +e^(−iθ) = 2 cos θ similarly e^(2iθ) +e^(−2iθ) = 2 cos 2θ ⇒ determinant (((cos 2θ = (1/2)(e^(2iθ) +e^(−2iθ) )))) also (i)−(ii) ⇒ e^(iθ) −e^(−iθ) = 2i sin θ similarly e^(2iθ) −e^(−2iθ) = 2i sin 2θ ⇒ determinant (((sin 2θ = (1/(2i))(e^(2iθ) −e^(−2iθ) )))) Hence Now sin^2 2θ = [(1/(2i))(e^(2iθ) −e^(−2iθ) )]^2 = −(1/4)(e^(4iθ) −2 + e^(−4iθ) ) cos^3 2θ = cos^2 2θ cos θ = [(1/2)(e^(2iθ) +e^(−2iθ) )]^2 [(1/2)(e^(2iθ) +e^(−2iθ) )] = (1/8)(e^(4iθ) +2+e^(−4iθ) )(e^(2iθ) +e^(−2iθ) ) cos^3 2θ = (1/8)(e^(6iθ) +3e^(2iθ) +3e^(−2iθ) +e^(−6iθ) ) ⇒ sin^2 2θ cos^3 2θ = −(1/(16))(e^(4iθ) −2+e^(−4iθ) )(e^(6iθ) +3e^(2iθ) +3e^(−2iθ) +e^(−6iθ) ) = −(1/(16))(e^(10iθ) +3e^(6iθ) +3e^(2iθ) +e^(−2iθ) −2e^(6iθ) −6e^(2iθ) −6e^(−2iθ) −2e^(−6iθ) +e^(2iθ) +3e^(2iθ) +3e^(−6iθ) +e^(−10iθ) ) = −(1/(16))(e^(10iθ) +10^(−10iθ) +e^(6iθ) +e^(−6iθ) +e^(2iθ) +e^(−2iθ) ) ⇒ 16 sin^2 2θ cos^3 2θ = 2 cos 2θ− cos 6θ−cos 10 θ as required. ∫_0 ^(π/3) 16 sin^2 2θ cos^3 2θ dθ = ∫_0 ^(π/3) (2 cos 2θ − cos 6θ − cos 10 θ) dθ = [sin2θ −((sin 6θ)/6)−((sin 10θ)/6)]_0 ^(π/3) = ((√3)/2)−0−0 ⇒ determinant (((∫_0 ^(π/3) 16 sin^2 2θ cos^3 2θ dθ = ((√3)/2))))](https://www.tinkutara.com/question/Q156254.png)
$${e}^{{i}\theta} =\:\mathrm{cos}\:\theta\:+\:{i}\:\mathrm{sin}\theta…….\left({i}\right) \\ $$$${e}^{−{i}\theta} \:=\:\mathrm{cos}\:\theta\:−\:{i}\:\mathrm{sin}\:\theta…..\left({ii}\right) \\ $$$$\mathrm{equation}\:\left({i}\right)\:\mathrm{and}\:\left({ii}\right)\:\mathrm{are}\:\mathrm{the}\:\mathrm{polar}\:\mathrm{and}\:\mathrm{index}\:\left(\mathrm{euler}\right)\:\mathrm{form} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{general}\:\mathrm{complex}\:\mathrm{number}\:\mathrm{with}\:\mathrm{modulus}\:\mathrm{1}. \\ $$$$\left({i}\right)\:+\:\left({ii}\right)\:\Rightarrow\:{e}^{{i}\theta} +{e}^{−{i}\theta} \:=\:\mathrm{2}\:\mathrm{cos}\:\theta\:\: \\ $$$$\mathrm{similarly}\:\:\:{e}^{\mathrm{2}{i}\theta} +{e}^{−\mathrm{2}{i}\theta} \:=\:\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\Rightarrow\:\begin{array}{|c|}{\mathrm{cos}\:\mathrm{2}\theta\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{\mathrm{2}{i}\theta} +{e}^{−\mathrm{2}{i}\theta} \right)}\\\hline\end{array} \\ $$$$\mathrm{also}\:\left({i}\right)−\left({ii}\right)\:\Rightarrow\:{e}^{{i}\theta} −{e}^{−{i}\theta} =\:\mathrm{2}{i}\:\mathrm{sin}\:\theta \\ $$$$\mathrm{similarly}\:\:{e}^{\mathrm{2}{i}\theta} −{e}^{−\mathrm{2}{i}\theta} =\:\mathrm{2}{i}\:\mathrm{sin}\:\mathrm{2}\theta \\ $$$$\Rightarrow\:\begin{array}{|c|}{\mathrm{sin}\:\mathrm{2}\theta\:=\:\frac{\mathrm{1}}{\mathrm{2}{i}}\left({e}^{\mathrm{2}{i}\theta} −{e}^{−\mathrm{2}{i}\theta} \right)}\\\hline\end{array} \\ $$$$\:\boldsymbol{\mathrm{Hence}} \\ $$$$\mathrm{Now}\:\mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta\:=\:\left[\frac{\mathrm{1}}{\mathrm{2}{i}}\left({e}^{\mathrm{2}{i}\theta} −{e}^{−\mathrm{2}{i}\theta} \right)\right]^{\mathrm{2}} \:=\:−\frac{\mathrm{1}}{\mathrm{4}}\left({e}^{\mathrm{4}{i}\theta} −\mathrm{2}\:+\:{e}^{−\mathrm{4}{i}\theta} \right) \\ $$$$\mathrm{cos}^{\mathrm{3}} \mathrm{2}\theta\:=\:\mathrm{cos}^{\mathrm{2}} \mathrm{2}\theta\:\mathrm{cos}\:\theta\:=\:\left[\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{\mathrm{2}{i}\theta} +{e}^{−\mathrm{2}{i}\theta} \right)\right]^{\mathrm{2}} \left[\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{\mathrm{2}{i}\theta} +{e}^{−\mathrm{2}{i}\theta} \right)\right]\:=\:\frac{\mathrm{1}}{\mathrm{8}}\left({e}^{\mathrm{4}{i}\theta} +\mathrm{2}+{e}^{−\mathrm{4}{i}\theta} \right)\left({e}^{\mathrm{2}{i}\theta} +{e}^{−\mathrm{2}{i}\theta} \right) \\ $$$$\mathrm{cos}^{\mathrm{3}} \mathrm{2}\theta\:=\:\frac{\mathrm{1}}{\mathrm{8}}\left({e}^{\mathrm{6}{i}\theta} +\mathrm{3}{e}^{\mathrm{2}{i}\theta} +\mathrm{3}{e}^{−\mathrm{2}{i}\theta} +{e}^{−\mathrm{6}{i}\theta} \right) \\ $$$$\Rightarrow\:\mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta\:\mathrm{cos}^{\mathrm{3}} \mathrm{2}\theta\:=\:−\frac{\mathrm{1}}{\mathrm{16}}\left({e}^{\mathrm{4}{i}\theta} −\mathrm{2}+{e}^{−\mathrm{4}{i}\theta} \right)\left({e}^{\mathrm{6}{i}\theta} +\mathrm{3}{e}^{\mathrm{2}{i}\theta} +\mathrm{3}{e}^{−\mathrm{2}{i}\theta} +{e}^{−\mathrm{6}{i}\theta} \right) \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{16}}\left({e}^{\mathrm{10}{i}\theta} +\mathrm{3}{e}^{\mathrm{6}{i}\theta} +\mathrm{3}{e}^{\mathrm{2}{i}\theta} +{e}^{−\mathrm{2}{i}\theta} −\mathrm{2}{e}^{\mathrm{6}{i}\theta} −\mathrm{6}{e}^{\mathrm{2}{i}\theta} −\mathrm{6}{e}^{−\mathrm{2}{i}\theta} −\mathrm{2}{e}^{−\mathrm{6}{i}\theta} +{e}^{\mathrm{2}{i}\theta} +\mathrm{3}{e}^{\mathrm{2}{i}\theta} +\mathrm{3}{e}^{−\mathrm{6}{i}\theta} +{e}^{−\mathrm{10}{i}\theta} \right) \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{16}}\left({e}^{\mathrm{10}{i}\theta} +\mathrm{10}^{−\mathrm{10}{i}\theta} +{e}^{\mathrm{6}{i}\theta} +{e}^{−\mathrm{6}{i}\theta} +{e}^{\mathrm{2}{i}\theta} +{e}^{−\mathrm{2}{i}\theta} \right) \\ $$$$\Rightarrow\:\mathrm{16}\:\mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta\:\mathrm{cos}^{\mathrm{3}} \mathrm{2}\theta\:=\:\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\theta−\:\mathrm{cos}\:\mathrm{6}\theta−\mathrm{cos}\:\mathrm{10}\:\theta\:\:\boldsymbol{\mathrm{as}}\:\boldsymbol{\mathrm{required}}. \\ $$$$ \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{3}}} {\int}}\mathrm{16}\:\mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta\:\mathrm{cos}^{\mathrm{3}} \mathrm{2}\theta\:{d}\theta\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{3}}} {\int}}\left(\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\theta\:−\:\mathrm{cos}\:\mathrm{6}\theta\:−\:\mathrm{cos}\:\mathrm{10}\:\theta\right)\:{d}\theta\:=\:\left[\mathrm{sin2}\theta\:−\frac{\mathrm{sin}\:\mathrm{6}\theta}{\mathrm{6}}−\frac{\mathrm{sin}\:\mathrm{10}\theta}{\mathrm{6}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \\ $$$$=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\mathrm{0}−\mathrm{0}\:\:\Rightarrow\:\begin{array}{|c|}{\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{3}}} {\int}}\mathrm{16}\:\mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta\:\mathrm{cos}^{\mathrm{3}} \mathrm{2}\theta\:{d}\theta\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\\\hline\end{array} \\ $$$$ \\ $$